Of course, since $\log i$ can become arbitrary large, so can $2^{\lceil\log i\rceil}$. However, I guess you meant something different. Maybe you want the maximal value of $2^{\lceil\log i\rceil}/i$?
–
celtschkSep 12 '12 at 16:28

If $i$ is $2^k+1$, $\lceil \log i \rceil = k+1$, so $2^{\lceil \log i \rceil} =2^{ k+1}$ and $2i-2^{\lceil \log i \rceil}=2i-2(i+1)=2$. If $i$ is further from a power of $2$, the expression will decline.

Can you explain why If $i$ is $2^k+1$, $\lceil \log i \rceil = k+1$
–
GeekSep 12 '12 at 16:33

1

The definition of $\log$ gives you that $k = \log 2^k$ whether k is integral or real. If you take $\log (2^k+1)$ for $k$ integral, it will be a little higher than $k$, so will round up to $k+1$
–
Ross MillikanSep 12 '12 at 16:37

1

@RossMillikan You have an extra factor of $2$ in $2(2i-2)$. The expression should always be positive since $\lceil \log i \rceil < \log i + 1$.
–
Erick WongSep 12 '12 at 17:29

To deal with $u(i)$ for $i$ integer, define $i_n$ as the unique integer such that $i_n\leqslant2^n\sqrt2\lt i_n+1$. Then $2i_n\gt2^{n+1}\sqrt2-2$ and $\lceil\log_2 i_n\rceil\leqslant n+1$ hence $u(i_n)\geqslant2^{n+1}(\sqrt2-1)-2$.

In particular, $u(i_n)\to+\infty$ when $n\to+\infty$, hence the family $\{u(i)\mid i\in\mathbb N\}$ is unbounded.

Likewise, for every real number $a\gt1$, the family $\{ai-a^{\lceil\log_ai\rceil}\mid i\in\mathbb N\}$ is unbounded.